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Tennis Ball dropping. Breaking a rack of pool balls Fastballs heating up a catcher’s glove. Glass of Water (freezing and evaporating). What does the First law of thermodynamics tell us about?. What happens to energy in each of these processes?
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Tennis Ball dropping. • Breaking a rack of pool balls • Fastballs heating up a catcher’s glove. • Glass of Water (freezing and evaporating) What does the First law of thermodynamics tell us about? What happens to energy in each of these processes? How is energy transferred from one form into another? It’s clear that (given the conditions) a process occurs spontaneously in one direction but not in another. What does the first law tell us about the reverse of each of these processes?
1st law of Thermodynamics • The first law came from the fact that we couldn’t create energy from nothing (it had to come from somewhere) • However, it places no limits on the conversion of energy from one type into another. • From the first law the possibility of any transformation of heat to work or work to heat exists as long as the amount of heat transferred exactly compensates for the amount of work. • The first law also makes no comment on whether a process will occur spontaneously or not.
The Second Law of Thermodynamics • The second law came from our observations that clearly there were limitations on what types of energy transfers happen spontaneously. • Heat does not move from a glove and zip the ball back to the pitcher. • A table of moving pool balls don’t organize themselves into a triangle and transfer all of their energy to the cue ball. • A tennis ball won’t start bouncing and end up in my hand. • Heat flows from a hot body to a cold body but not vice versa. • All the molecules in the air in this room do not simultaneously decide to all move in the same direction (and do work). • When it’s freezing outside the water in a glass freezes when it’s hot and dry, it evaporates. • What do each of these processes have in common? lete this box.: AAAAAAAAAx
Dispersal of Energy is Spontaneous The forward path (spontaneous path) of each of these processes involves the dispersal of energy (spontaneous) The reverse or concentration of energy is not spontaneous. Another piece of evidence for the second law was that we couldn’t build a perpetual motion machine. If we put a certain amount of energy in the machine to the right the energy would eventually spontaneously disperse. lete this box.: AAAAAAAAAx
The Second Law According to Lord Kelvin and Clausius A transformation whose only final result is to transform into work heat extracted from a source which is at the same temperature is impossible (Postulate of Lord Kelvin ) If heat flows by conduction from a body A to another body B, then a transformation whose only final result is to transfer heat from B to A is impossible. Postulate of Clausius (These two happen to be equivalent). Let’s prove it. TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAAAAAAAA
2nd law • Let’s show that Kelvin’s and Clausius’ statements are equivalent. • We will do this by showing that a violation of Kelvin’s statement violates Clausius’ statement and that a violation of Clausius’ statement violates Kelvin’s statement.
Before we go on, a reminder about one of Joule’s Experiments Joule’s experiment showed that Work can be directly transformed into an equal amount of heat. First, let’s assume that Kelvin’s postulate is not valid. Then we could extract heat from a cooler source Tc and convert it into entirely to work then take that work and through friction heat up Th (Joule’s experiment showed that work could be entirely converted to heat) The net result would be having heat flow from a cold source at Tc to a hot source Th Before we do the second half of the proof, let’s talk about the possibilities of transforming heat into work.
The Carnot cycle Carnot Cycle Named for Sadi Carnot (1796- 1832) Isothermal expansion Adiabatic expansion Isothermal compression Adiabatic compression
a • Qh b • Th Q=0 Q=0 • d • Tc Qc c Carnot Engine (heat engine) Pressure Volume
a • Qh b • Th W Q=0 q=0 • d • Tc Qc c Carnot Cycle Pressure Volume
a • b • • d c • What are the signs of q and w for each of the steps in the Carnot Process? How is the system coupled to the surroundings in each step? How is the temperature maintained during the isothermal steps?
2nd Law Clausius according to Kelvin • Allow an amount of heat Qc to flow from reservoir Tc to reservoir Th • Then we would do a carnot cycle operating between Tc and Th in which Qc is absorbed by Tc from the system. • The net effect is that the reservoir Tc and the system are left unchanged (an equal amount of heat flowed into and out of it), work has been done and heat has been exchanged between the system and reservoir Th • This would violate Kelvin’s statement of the 2nd law. Heat has transferred from a source (Th) at uniform Temperature and work has been done • Therefore Clausius’ and Kelvin’s statements are equivalent.
The Carnot cycle for a gas might occur as visualized below. Hot reservoir th at Th Cool reservoir tc at Tc QL
Carnot Engine If we couple the piston to a crankshaft to a connecting rod, we can drive a car. A two stroke engine
Carnot Cycle For Any Cycle, all of the state variables remain unchanged since, for the whole process, there is no change of state. During a clockwise cycle, the system gives up Qc to heat reservoir Tc and absorbs Qh from heat reservoir Th then the total work and heat are: We define the efficiency of the engine as the amount of work done divided by the heat absorbed by the system.:
Carnot Cycle From a to b: isothermal, so that DU = 0 and q = - W Thus, Qh = +nRThln(Vb/Va) From b to c: adiabatic, Q = 0, so that TVg-1 is constant. Thus, ThVbg-1 = TcVcg-1 or Similarly, from c to d: isothermal, so that DU = 0 and q = - w Thus, Qc = +nRTcln(Vd/Vc) = -nRTcln(Vc/Vd) Similarly, d to a: adiabatic, Q = 0, so that TVg-1 is constant. Thus, TcVdg-1 = ThVag-1 or
Which means that Carnot Cycle We see that: Now also from the isothermal parts of the carnot cycle: This is an important result. Temperature can be defined (on the absolute (Kelvin) scale) in terms of the heat flows in a Carnot Cycle. This is important since it works for any working fluid and does not depend on the properties of an Ideal Gas
Temperature and its scales • Constant-volume gas thermometer and the Kelvin scale • A constant-volume gas thermometer measures the pressure of • the gas contained in the flask immersed in the bath. The volume • of the gas in the flask is kept • constant by raising or lowering • reservoir B to keep the mercury • level constant in reservoir A.
Absolute Zero and the Kelvin Scale With ice being zero and boiling water being 100 the heights from The constant volume thermometer could be gradated into degrees of Celsius. That’s how the scale came into being. Extrapolation of measurements made using different gases leads to the concept of absolute zero, when the pressure (or volume) is zero.
Carnot Cycle We can also relate the efficiency of the engine to the ratio of the temperatures of the two sources. From this alone, we can get ideas on how to make our engines more efficient? How? (Think of engine efficiency on a hot day and engine efficiency on a cold day)
Carnot Cycle If the system does work on the surroundings. That is that w is negative then we can show that the system surrendered heat to the lower temperature source and received heat from the higher temperature source. Proof (Part 1): Let’s say that the system received heat from the cooler source then we could put Tc and Th in thermal contact until Qc heat flowed from Th to Tc. This would leave us with work done extracted from a single source with the same temperature throughout which is in violation of Kelvin’s postulate. So, heat is surrendered from the system to the cooler source. Qc < 0 Proof (Part 2): The second part is now easy since work is negative then Qh must also be negative since -Qc is positive